Question

The Hubble Space Telescope orbits the Earth at an approximate altitude of 612 km. Its mass is 11,100 kg and the mass of the Earth is 5.97×1024 kg. The Earth's average radius is 6.38×106 m. What is the magnitude of the gravitational force that the Earth exerts on the Hubble?

Answer 1

Answer:

magnitude of the gravitational force is 9.04 × $10^{4}$ N

Explanation:

given data

altitude = A = 612 km = 612000 m

mass M = 11,100 kg

mass of the Earth m = 5.97 × $10^{24}$ kg

Earth average radius = 6.38 × $10^{6}$  m

to find out

magnitude of the gravitational force

solution

first we get here distance from space to centre of earth that is

distance = altitude + earth radius

distance = 612000  +  6.38 × $10^{6}$  m

distance = 6.99 × $10^{6}$  m  

so now we get here  magnitude of the gravitational force that is express as

magnitude of the gravitational force F = $\frac{G*M*m}{distance^2}$   ...........1

here G is gravitational constant  that is 6.67 × $10^{-11}$ Nm² /kg and M is mass of space and m is mass of earth

put here all value we get

F = $\frac{G*M*m}{distance^2}$

F = $\frac{6.67*10^{-11}*5.97*10^{24}*11100}{(6.99*10^{6})^2}$

F = 9.04 × $10^{4}$ N

so magnitude of the gravitational force is 9.04 × $10^{4}$ N